Inserting (REF ) into the non-conservative compressible Euler system (REF ) to yield
\(\begin{aligned}&\partial _t\left\lbrace \delta \left(\overline{E}_0 \sigma +\left(3\overline{E}_2/2-\overline{K}_g \overline{E}_0\right)\theta \right)+\delta ^2 \rho _d^\delta \right\rbrace \\&\qquad +(\delta u+\delta ^2 u_d^\delta )\cdot \nabla _x\left\lbrace \delta \left(\overline{E}_0 \sigma +\left(3\overline{E}_2/2-\overline{K}_g \overline{E}_0\right)\theta \right)+\delta ^2 \rho _d^\delta \right\rbrace \\&\qquad \qquad +\left\lbrace \overline{\rho }+\delta \left(\overline{E}_0 \sigma +\left(3\overline{E}_2/2-\overline{K}_g \overline{E}_0\right)\theta \right)+\delta ^2 \rho _d^\delta \right\rbrace \nabla _x\cdot (\delta u+\delta ^2 u_d^\delta )=0\\&\left\lbrace \overline{\rho }+\delta \left(\overline{E}_0 \sigma +\left(3\overline{E}_2/2-\overline{K}_g \overline{E}_0\right)\theta \right)+\delta ^2 \rho _d^\delta \right\rbrace \partial _t(\delta u+\delta ^2 u_d^\delta )\\&\qquad +\left\lbrace \overline{\rho }+\delta \left(\overline{E}_0 \sigma +\left(3\overline{E}_2/2-\overline{K}_g \overline{E}_0\right)\theta \right)+\delta ^2 \rho _d^\delta \right\rbrace (\delta u+\delta ^2 u_d^\delta )\cdot \nabla _x(\delta u+\delta ^2 u_d^\delta )\\&\qquad \qquad + \nabla _x(\delta \overline{E}_2(\sigma +\theta )+\delta ^2 \mathcal {E}_d^\delta )=0\\&\partial _t (\delta \overline{E}_2(\sigma +\theta )+\delta ^2 \mathcal {E}_d^\delta )+(\delta u+\delta ^2 u_d^\delta )\cdot \nabla _x(\delta \overline{E}_2(\sigma +\theta )+\delta ^2 \mathcal {E}_d^\delta )\\&\qquad \qquad + \frac{5}{3}(\overline{\mathcal {E}}+\delta \overline{E}_2(\sigma +\theta )+\delta ^2 \mathcal {E}_d^\delta )\nabla _x\cdot (\delta u+\delta ^2 u_d^\delta )=0.\end{aligned}\) 
Noting that \((\sigma , u, \theta )\)  satisfy the acoustic system (REF ). We can deduce that the coefficients of \(\delta \)  in the equations above are all equal to 0:
\(\left\lbrace \begin{aligned}&\partial _t\left\lbrace \left(\overline{E}_0 \sigma +\left(3\overline{E}_2/2-\overline{K}_g \overline{E}_0\right)\theta \right)\right\rbrace +\overline{\rho }\nabla _x\cdot u=0,\\&\overline{\rho }\partial _t u +\overline{E}_2\nabla _x (\sigma +\theta )=0,\\&\overline{E}_2\partial _t (\sigma +\theta )+\frac{5}{3}\overline{\mathcal {E}}\nabla _x\cdot u=0,\end{aligned}\right.\) 
since we have \(\overline{\rho }=\overline{E}_2\)  and \(\overline{K}_A=\frac{5\overline{\mathcal {E}}}{2\overline{\rho }}\) . Therefore the remaining terms which are the coefficients of \(\delta ^2\)  form the equations:
\(\begin{aligned}&\partial _t\rho _d^\delta +\mathfrak {u}^\delta \cdot \nabla _x \rho _d^\delta +\rho ^\delta \nabla _x\cdot u_d^\delta +\delta \left\lbrace u_d^\delta \cdot \nabla _x\left[\overline{E}_0 \sigma +\left(3\overline{E}_2/2-\overline{K}_g \overline{E}_0\right)\theta \right]+\rho _d^\delta \nabla _x\cdot u\right\rbrace \\&\hspace{199.16928pt}=-\nabla _x\cdot \Big \lbrace \left[\,\overline{E}_0 \sigma +\left(3\overline{E}_2/2-\overline{K}_g \overline{E}_0\right)\theta \,\right]u\Big \rbrace \\&\rho ^\delta \partial _t u_d^\delta +\rho ^\delta \mathfrak {u}^\delta \cdot \nabla _x u_d^\delta +\nabla _x\mathcal {E}_d^\delta +\delta \left( \rho _d^\delta \partial _t u+\rho ^\delta u_d^\delta \cdot \nabla _x u \right) \\&\hspace{199.16928pt}=-\left[\overline{E}_0 \sigma +\left(3\overline{E}_2/2-\overline{K}_g \overline{E}_0\right)\theta \right]\partial _t u-\rho ^\delta u\cdot \nabla _x u\\&\partial _t \mathcal {E}_d^\delta +\mathfrak {u}^\delta \cdot \nabla _x\mathcal {E}_d^\delta +\frac{5}{3}\mathcal {E}^\delta \nabla _x\cdot u_d^\delta +\delta \left[\overline{E}_2 u_d^\delta \cdot \nabla _x(\sigma +\theta )+\frac{5}{3}\mathcal {E}_d^\delta \nabla _x\cdot u\right]\\&\hspace{199.16928pt}=-\overline{E}_2 u\cdot (\sigma +\theta )-\frac{5}{3}\overline{E}_2\nabla _x(\sigma +\theta )\nabla _x\cdot u.\end{aligned}\) 
Indeed, the equations above are the system governed by the difference variables \((\rho _d^\delta ,u_d^\delta ,\mathcal {E}_d^\delta )\)  with the coefficients depending on the smooth Euler solution \((\rho ^\delta ,\mathfrak {u}^\delta ,\mathcal {E}^\delta )\)  and smooth acoustic solution \((\sigma ,u,\theta )\)  over time integral \([0,\tau ]\) . Then going a step further, we can formulate (REF ) as a symmetric hyperbolic system:
\(A_0^\delta \partial _t U_d +\sum _{i=1}^{3}A_i^\delta \partial _i U_d +B^\delta U_d = F,\) 
where \(A_0^\delta \) , \(A_i^\delta \) , \(B^\delta \)  and \(F\)  are given by
\(U_d=\left(\begin{aligned}\mathcal {E}_d^\delta \\ \mathfrak {u}_d^\delta \\ \rho _d^\delta \end{aligned}\right),\quad A_0^\delta =\left(\begin{array}{ccc} \frac{6}{5}& 0 &-\frac{\mathcal {E}^\delta }{\rho ^\delta }\\ 0 & \rho ^\delta \mathcal {E}^\delta \mathbb {I} & 0 \\ -\frac{\mathcal {E}^\delta }{\rho ^\delta } & 0 & \frac{5(\mathcal {E}^\delta )^2}{3(\rho ^\delta )^2}\end{array}\right),\quad A_i^\delta =\left(\begin{array}{ccc} \frac{6}{5}\mathfrak {u}_i^\delta & \mathcal {E}^\delta e_i & -\frac{\mathcal {E}^\delta }{\rho ^\delta }\mathfrak {u}_i^\delta \\ \mathcal {E}^\delta (e_i)^\mathrm {T} & \rho ^\delta \mathcal {E}^\delta \mathfrak {u}_i^\delta \mathbb {I} & 0 \\ -\frac{\mathcal {E}^\delta }{\rho ^\delta }\mathfrak {u}_i^\delta & 0 & \frac{5(\mathcal {E}^\delta )^2}{3(\rho ^\delta )^2}\mathfrak {u}_i^\delta \end{array}\right),\) 
\(B^\delta =\delta A_0^\delta \left(\begin{array}{ccc}\frac{5}{3}\nabla _x\cdot u & \overline{E}_2 \nabla _x(\sigma +\theta ) & 0 \\ 0 & \rho ^\delta \nabla _x u & \partial _t u\\ 0 & \nabla _x\left[\overline{E}_0 \sigma +\left(3\overline{E}_2/2-\overline{K}_g \overline{E}_0\right)\theta \right] & \nabla _x\cdot u \end{array}\right),\) 
\(F=-A_0^\delta \left(\begin{aligned}\overline{E}_2 u\cdot (\sigma +\theta )+\frac{5}{3}\overline{E}_2\nabla _x(\sigma +\theta )\nabla _x\cdot u\\\left[\overline{E}_0 \sigma +\left(3\overline{E}_2/2-\overline{K}_g \overline{E}_0\right)\theta \right]\partial _t u+\rho ^\delta u\cdot \nabla _x u \\\nabla _x\cdot \Big \lbrace \left[\,\overline{E}_0 \sigma +\left(3\overline{E}_2/2-\overline{K}_g \overline{E}_0\right)\theta \,\right]u\Big \rbrace \end{aligned}\right).\) 
Here, \(\mathbb {I}\)  is the \(3\times 3\)  identity matrix and \(e_i\) 's \((i=1,2,3)\)  are the standard unit row base vectors in \(\mathbb {R}^3\) . \((\cdot )^\mathrm {T}\)  denotes the transpose of row vectors. Similar to the arguments in section , the system (REF ) is strictly hyperbolic. Hence by the Friedrichs’ existence theory of the linear hyperbolic system (c.f. Chapter 2 in [1]}), the existence of smooth solution to the system (REF ) is guaranteed and in addition for \(s\geqslant 3\) , we have the following energy inequality:
\(\frac{\mathrm {d}}{\mathrm {d}t}\Vert U_d\Vert _{H^s}^2 \leqslant C_3\Vert U_d\Vert _{H^s}^2+C_4 \Vert U_d\Vert _{H^s},\) 
where \(C_3\)  and \(C_4\)  are depending on the \(H^{s+1}\)  norms of \((\rho ^\delta ,\mathfrak {u}^\delta ,\mathcal {E}^\delta )\)  and \((\sigma ,u,\theta )\) . A simple Grönwall inequality combining with the estimates (REF ) implies that \(\Vert (\rho _d^\delta ,u_d^\delta ,\mathcal {E}_d^\delta )\Vert _{H^s}\)  is bounded by a constant that depends only on \(\tau \)  and \(\Vert (\sigma ^\mathrm {in},u^\mathrm {in},\theta ^\mathrm {in})\Vert _{H^{s+1}}\) . We obtain the conclusion as announced.
